Question

What will be the de – Broglie wavelength when the kinetic energy of the electron increases by 5 times?

(a) √5

(b) 5

(c) \(\frac {1}{\sqrt {5}}\)

(d) \(\frac {1}{5}\)

I have been asked this question in an interview for internship.

I want to ask this question from Wave Nature of Matter topic in portion Dual Nature of Radiation and Matter of Physics – Class 12

Answer 1

Correct answer is (c) \(\frac {1}{\sqrt {5}}\)

To explain: The required equation ➔ λ=\(\frac {h}{mv} = \frac {h}{\sqrt {2mK}} \)

Where h is the Planck’s constant, m is the mass of the electron and K is the kinetic energy of the electron.

Since the mass of the electron remain unchanged, the wavelength will be inversely proportioned to the kinetic energy, so,

\(\frac {\lambda }{\lambda^{‘}} = \sqrt {\frac {K^{‘}}{K} } = \sqrt { \frac {5K}{K} } \) = √5

Therefore, λ’=\(\frac {\lambda }{\sqrt {5}}\)

Hence, the wavelength is reduced by a factor of √5.